Ian Thompson-Bell wrote:
>
> I finally cracked it - or rather found the text with an explanation I
> could understand. Basically, with shunt applied feedback it is currents
> that are summed/subtracted at the input. As the output is a voltage the
> stage is a transconductance amplifier so the closed loop gain is in the
> form of Vo/Ii which does turn out to be Ao/(1+ß.Ao) and ß is simply 1/Rf
> where Rf is the feedback resistor
Indeed the the transconductance amp is one with extremely high input
resistance.
So there is no input current to the amp, so a tube grid or fet gate
input
works well with shunt FB.
For ß to be equal to 1/Rf, then how come? where does the 1 come from?
What is it?
How come you don't express ß as Ri / ( Ri + Rf) , or in my way, R1 / (
R1 + R2 ) ?
The Ii or input current supplied by the voltage signal source flows
equally in R1 and R2.
But you have to know what the grid voltage is and Vo to sum these to get
the current in R2.
Or if you like, Ii = ( A + 1 ) / R2, where +A volts is at output and -1
volt is at the grid.
So the voltage across R1 = R1 x [ ( A + 1 ) / R2 ] = ( R1/R2 ) x ( A + 1
)
To get the input voltage you must add the R1 voltage to the grid voltage
which is 1, so you get
Vin = ( R1/R2 ) x ( A + 1 ) + 1.
closed loop A' = Vo / Vi. But Vo = +A volts, with -1V at the grid.
So A' = A / [ ( R1/R2 ) x ( A+1 ) + 1 ]
Consider a 1/2 6SN7, A = 16, R1 = 40k, R2 = 80k.
Say = +16 is at anode, then -1V is at grid, so voltage across 80k = 17V,
and across 40k is 8.5V, so input voltage MUST be -9.5V, ok?
A' = 16 / 9.5 .
Using the formula above without ß in it, just R1 and R2,
A' = 16 / [ 0.5 x (16 + 1) + 1 ] = 16 / 9.5
You should be able to prove whether your reasons and formulas work
with common types of triodes or pentodes,
and make sure other dumb folks will understand you.
>
> The input resistor effectively inputs a current equal to Vi/Ri from
> which you can work out the voltage gain as Rf/Ri since the closed loop
> transconductance is Rf. I'll include this in NFB101.
The input current is not Vin / R1.
At the grid there is a smaller signal of the same phase as the input
signal.
So the current in R1 = ( Vin - Vg ) / R1.
Patrick Turner.
>
> Cheers
>
> Ian
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